Question: For $1\leq i, k \leq n$ you are given some real numbers $b_i$ and $c_{ik}$ so that:
$$\sum_{i,k=1}^{n} c^2_{ik} < 1$$
Show, using Banach fixed-point theorem, that the following non-linear system of equations has exactly one solution:
$$x_i = \sum^{n}_{k=1} \sin (c_{ik}x_k) + b_i, \qquad 1 \leq i \leq n$$
My progress: I have to show, that $$d(f(x),f(y)) < q \cdot d(x,y), \quad 0 <q< 1$$
I took the equation and tried to play around with it:
$$\begin{aligned}
d(f(x), f(y))^2
&= \Big(\sum^{n}_{k=1}\sin(c_{1k}x_k)-\sum^{n}_{k=1}\sin(c_{1k}y_k)\Big)^2 + …+
\Big(\sum^{n}_{k=1}\sin(c_{nk}x_k)-\sum^{n}_{k=1}\sin(c_{nk}y_k)\Big)^2 \\
&= \Big(\sum^{n}_{k=1}\big(\sin(c_{1k}x_k)-\sin(c_{1k}y_k)\big)\Big)^2+…+\sum^{n}_{k=1}\big(\sin(c_{nk}x_k)-\sin(c_{nk}y_k)\big)\Big)^2 \\
&= \bigg(\sum^{n}_{k=1}2\sin\Big(\frac{c_{1k}(x_k-y_k)}{2}\Big)\bigg)^2 + … + \bigg(\sum^{n}_{k=1}2\sin\Big(\frac{c_{nk}(x_k-y_k)}{2}\Big)\bigg)^2\end{aligned}$$
Then I could use $(z_1+z_2+…+z_n)^2 \leq n \sum c_{ik} (z^2_1+z^2_2+…+z^2_n)$ to show that it contracts, since $\sum c_{ik} <1$, but I can't evaluate it without knowing what $n$ is… Feel out of options. Any help strongly appreciated!
Best Answer
Let $\phi:\mathbb{R}^n \to \mathbb{R}^n$ be given by $\phi(x)_i= \sum^{n}_{k=1} \sin (c_{ik}x_k) + b_i$.
\begin{eqnarray} |\phi(x)_i - \phi(y)_i| &=& \sum^{n}_{k=1} |\sin (c_{ik}x_k) -\sin (c_{ik}y_k)| \\ &\le& \sum^{n}_{k=1} |c_{ik}||x_k-y_k| \\ &\le& \sqrt{\sum^{n}_{k=1} c_{ik}^2}\|x-y\| \end{eqnarray} I used the mean value theorem and Cauchy Schwartz above.
Hence $\|\phi(x)-\phi(y)\| \le \sqrt{\sum^{n}_{i,k=1} c_{ik}^2 } \|x-y\| $